如何按类获取图像src

Question

我有这个:

<a href="/Dealer-Catalog/ManufacturerID-3"><img class="brand-logo" src="http://www.teledynamics.com/tdresources/74c42cb2-dc7f-4548-b820-2946fbe160db.jpg" onerror="this.src='/Content/Css/Images/no_brand_logo_120_48.gif'" alt="ADTRAN"></a>

如何获得img src(http://www.teledynamics.com/tdresources/74c42cb2-dc7f-4548-b820-2946fbe160db.jpg)

我尝试了很多认为这是最后一个:

$doc = new DOMDocument();
libxml_use_internal_errors(true);
$doc->loadHTML($html);
$xpath = new DOMXPath($doc);
$src = $xpath->evaluate("string(//class='brand-logo']/img/@src)");
echo "$src";

Answer 1

这不是正确的XPath语法.尝试

$nodes = $xpath->query("//img[@class='brand-logo']");
$src = $nodes->item(0)->getAttribute('src');

首先,您获取代表您想要的src图像的NODE,然后获取src属性.请注意, - > query()调用返回DOMNodeList,而不是节点.